I've tried to compare it to the series [itex]\sum_{i=1}^{\infty} \frac{\sqrt{2n-1} \log{(4n + 1)}}{n^2}[/itex] and show the latter converges. I have no idea how to show this. Although the limit of the sequence approaches 0 as n goes to infinite, that is not enough to guarantee convergence. The book says it converges.

I've tried to compare it to the series [itex]\sum_{i=1}^{\infty} \frac{\sqrt{2n-1} \log{(4n + 1)}}{n^2}[/itex] and show the latter converges. I have no idea how to show this. Although the limit of the sequence approaches 0 as n goes to infinite, that is not enough to guarantee convergence. The book says it converges.

Let t(n) = nth term above. You could try to get a simple upper bound on t(n): sqrt(2n-1) < sqrt(2n), log(4n-1) < log(4n) and n(n+1) > n^2. Thus, t(n) < sqrt(2n)*log(4n)/n^2, which is of the form c*log(n)/n^(3/2). Convergence of sum log(n)/n^(3/2) is easier to show, and that implies convergence of sum t(n) [why?]